use the exterior angle inequality theorem to the exterior angle inequality theorem to list all of...

Use the Exterior Angle Inequality Theorem to list all of the angles that satisfy the stated condition.

1. measures less than

SOLUTION: By the Exterior Angle Inequality Theorem, the

exterior angle ( ) is larger than either remote

interior angle ( and ). Also, , and

. By substitution,

. Therefore, must be larger than each individual angle. By the Exterior Angle Inequality Theorem,

2. measures greater than m 7

exterior angle ( 5) is larger than either remote

interior angle ( 7 and 8). Similarly, the exterior

angle ( 9) is larger than either remote interior angle

( 7 and 6). Therefore, 5 and 9 are both largerthan .

interior angle ( 1 and 2). Therefore, 4 is larger than .

4. measures less than m 9

interior angle ( and ). Also, , and (

. By substitution, . Therefore, must be larger than each individual

angle so is larger than both and .

List the angles and sides of each triangle in order from smallest to largest.

SOLUTION:

Here it is given that, . Therefore, from angle-side relationships in a triangle, we know that Angle:

SOLUTION: By the Triangle Sum Theorem,

So, From angle-side relationships in a triangle, we know

Angle: J, K, L

7. HANG GLIDING The supports on a hang glider form triangles like the one shown. Which is longer —

the support represented by or the support

represented by ? Explain your reasoning.

SOLUTION:

Since the angle across from segment is larger

than the angle across from , we can conclude that

is longer than due to Theorem 5.9, which states that if one side of a triangle is longer than another, then the angle opposite the longer side has a greater measure than the angle opposite the shorter side.

CCSS SENSE-MAKING Use the Exterior Angle Inequality Theorem to list all of the angles that satisfy the stated condition.

interior angle ( 1 and 2). Therefore, m 4 >

exterior angle ( 4) is greater than either remote

interior angle ( 1 and 2).Therefore, m 1 < m 4 and m 2 < m 4.

interior angle ( 7 and 8). Therefore, m 7 <

m 5 and m 8 < m 5 .

interior angle from two different triangles ( 6 and

7 from one triangle and and from another. Therefore, , ,

, and .

( 7 and 8). Therefore, m 2 > m 8 and m 5

> m 8 .

SOLUTION: The hypotenuse of the right triangle must be greater than the other two sides.

Therefore, By Theorem 5.9, the measure of the angle opposite the longer side has a greater measure tahn the angle opposite the shorter

side, therefore Angle:

SOLUTION:

Based on the diagram, we see that By Theorem 5.9, the measure of the angle opposite the longer side has a greater measure than the angle opposite the shorter side, therefore

Angle:

SOLUTION:

Therefore, by Theorem 5.10 we know that the side opposite the greater angleis longer than the side opposite a lesser angle and

Angle:

SOLUTION:

. Therefore, by Theorem5.10 we know that the side opposite the greater angleis longer than the side opposite a lesser angle and

Angle:

20. SPORTS Ben, Gilberto, and Hannah are playing Ultimate. Hannah is trying to decide if she should pass to Ben or Gilberto. Which player should she choose in order to have the shorter passing distance?Explain your reasoning.

SOLUTION: Based on the Triangle Sum Theorem, the measure ofthe missing angle next to Ben is

degrees. Since 48 < 70, theside connecting Hannah to Ben is the shortest, based on Theorem 5.10. Therefore, Hannah should pass to Ben.

21. RAMPS The wedge below represents a bike ramp.

Which is longer, the length of the ramp or the

length of the top surface of the ramp ? Explain your reasoning using Theorem 7.9.

SOLUTION:

If m X = 90, then , based on the Triangle Sum

Theorem, m Y + m Z = 90, so m Y < 90 by the

definition of inequality. So m X > m Y. According to Theorem 5.10, if m X > m Y, then

the length of the side opposite X must be greater

than the length of the side opposite Y. Since is

opposite X, and is opposite Y, then YZ > XZ.Therefore YZ, the length of the top surface of the ramp, must be greater than the length of the ramp.

SOLUTION: Using the Triangle Sum Theorem, we can solve for x, as shown below.

degrees and the

degrees. Therefore,

. By Theorem 5.10, we know that the lengths of sides across from larger angles are longer than those across from shorter angles so

. Angle:

degrees, degrees and the

degrees. Therefore,

. By Theorem 5.10, we know that the lengths of sides across from larger angles are longer than those across from shorter

angles so . Angle: P, Q, M

Use the figure to determine which angle has thegreatest measure.

24. 1, 5, 6

SOLUTION: By the Exterior Angle Inequality, we know that the measure of the exterior angle of a triangle is greater than the measure of either of its corresponding remote interior angles. Therefore, since is the exterior angle and and are its remote interior

angles, the m 1 is greater than m 5 and m 6.

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

SOLUTION: By the Exterior Angle Inequality, we know that the measure of the exterior angle of a triangle is greater than the measure of either of its corresponding remote interior angles. Therefore, since is the exterior angle and and are its remote

interior angles, the m 3 is greater than m 11 and

28. 3, 9, 14

SOLUTION: By the Exterior Angle Inequality, we know that the measure of the exterior angle of a triangle is greater than the measure of either of its corresponding remote interior angles. Therefore, since is the exterior angle and and the sum of and

are its remote interior angles, the m 3 is greater

than m 9 and the sum of the m 11 and m 14. If is greater than the sum of and ,

then it is greater than the measure of the individual

angles. Therefore, the m 3 is greater than m 9

and m 14.

29. 8, 10, 11

are its remote interior angles, the is greater than m 10 and the sum of the m 11 and

m 14. If is greater than the sum of and, then it is greater than the measure of the

individual angles. Therefore, the m 8 is greater than

m 10 and m 11.

CCSS SENSE-MAKING Use the figure to determine the relationship between the measures of the given angles.

30. ABD, BDA

SOLUTION:

The side opposite is , which is of length 13.

Since by Theorem 5.9.

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

The side opposite is , which is of length

33. DBF, BFD

SOLUTION:

Use the figure to determine the relationship between the lengths of the given sides.

34. SM, MR

SOLUTION:

Since are a linear pair,

The side opposite is . The side opposite is . In

, , since 60 < 70.

Therefore, by Theorem 5.10, .

35. RP, MP

SOLUTION:

Since form a straight angle, . The side opposite is . The side opposite is . In ,

, since 70 < 35. Therefore,

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

The side opposite is . The side opposite is . In ,

, since 30 < 85. Therefore, by

Theorem 5.10,

37. RM, RQ

. The side

opposite ∠PQR is . The side opposite ∠RPQ is

. In , m∠PQR > m∠RPQ, since 65 > 30.Therefore, by Theorem 5.10, PR > RQ. Also, form a straight angle,

and, by the Triangle Sum Theorem,

. The side opposite ∠MPR is . The side opposite ∠PMR is . In , m∠MPR >

m∠PMR, since 75 > 70. Therefore, by Theorem

5.10, . If and , then by the transitive

property of inequality, Thus,

38. HIKING Justin and his family are hiking around a lake as shown in the diagram. Order the angles of the triangle formed by their path from largest to smallest.

SOLUTION: The side opposite is of length 0.5. The side opposite is of length 0.4.

The side opposite is of length 0.45.

Since 0.5 > 0.45 > 0.4, by Theorem 5.9, m 3 >

m 1 > m 2.

COORDINATE GEOMETRY List the angles of each triangle with the given vertices in order from smallest to largest. Justify your answer.

39. A(–4, 6), B(–2, 1), C(5, 6)

SOLUTION: Use the Distance Formula to find the side lengths.

40. X(–3, –2), Y(3, 2), Z(–3, –6)

41. List the side lengths of the triangles in the figure fromshortest to longest. Explain your reasoning.

SOLUTION: AB, BC, AC, CD, BD; In , AB < BC < AC and in , BC < CD < BD. By the figure AC < CD, so BC < AC < CD.

42. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between the sides of a triangle.

a. GEOMETRIC Draw three triangles, including one acute, one obtuse, and one right angle. Label the vertices of each triangle A,B, and C. b. TABULAR Measure the length of each side of the three triangles. Then copy and complete the table. c. TABULAR Create two additional tables like the one above, finding the sum of BC and CA in one table and the sum of AB and CA in the other. d. ALGEBRAIC Write an inequality for each of thetables you created relating the measure of the sum of two of the sides to the measure of the third side of a triangle. e. VERBAL Make a conjecture about the relationship between the measure of the sum of two sides of a triangle and the measure of the third side.

SOLUTION: a.

b. It might be easier to use centimeters to measure the sides of your triangles and round to the nearest tenth place, as shown in the table below. Sample answer:

c. It might be easier to use centimeters to measure the sides of your triangles and round to the nearest

tenth place, as shown in the table below. Sample answer:

d. Compare the last two columns in each table and write an inequality comparing each of them. AB + BC > CA, BC + CA > AB, AB + CA > BC e . Sample answer: The sum of the measures of two sides of a triangle is always greater than the measureof the third side of the triangle.

43. WRITING IN MATH Analyze the information given in the diagram and explain why the markings must be incorrect.

SOLUTION:

Sample answer: R is an exterior angle to ,

so, by the Exterior Angle Inequality, m R must be

greater than m Q, one of its corresponding remote

interior angles. The markings indicate that ,

indicating that m R = m Q. This is a contradiction

of the Exterior Angle Inequality Theorem since

can't be both equal to and greater than at the same time. Therefore, the markings are incorrect.

44. CHALLENGE Using only a ruler, draw

such that m A > m B > m C. Justify your drawing.

SOLUTION: Due to the given information, that

, we know that this is a triangle with three different angle measures. Therefore, it also has three different side lengths. You might want to start your construction with the longest side, which would be across from the largest

angle measure. Since A is the largest angle, the

side opposite it, , is the longest side. Then, make ashort side across from the smallest angle measure.

Since C is the smallest angle, is the shortest side. Connect points A to C and verify that the length

of is between the other two sides.

45. OPEN ENDED Give a possible measure for in shown. Explain your reasoning.

SOLUTION:

Since m C > m B, we know, according to

Theorem 5.10, that the side opposite must

be greater than the side opposite . Therefore, if AB > AC, then AB > 6 so you need to choose a length for AB that is greater than 6. Sample answer: 10; m C > m B so AB > AC, Therefore, Theorem 5.10 is satisfied since 10 > 6.

46. CCSS ARGUMENTS Is the base of an isosceles triangle sometimes, always, or never the longest sideof the triangle? Explain.

SOLUTION: To reason through this answer, see if you can sketch isosceles triangles that have a base shorter than the two congruent legs, as well as longer than the two congruent legs. Sometimes; Sample answer: If the measures of the base angles are less than 60 degrees, then the base will be the longest leg. If the measures of the base angles are greater than 60 degrees, then the base will be the shortest leg.

47. CHALLENGE Use the side lengths in the figure to list the numbered angles in order from smallest to

largest given that m 2 = m 5. Explain your reasoning.

SOLUTION: Walk through these angles one side at a time. Given: m 2 = m 5 The side opposite 5 is the smallest side in that

triangle and m 2 = m 5, so we know that m 4

and m 6 are both greater than m 2 and m 5.

Also, the side opposite m 6 is greater than the side

opposite m 4. So far, we have (m 2 = m 5) < m 4 < m 6. Since the side opposite 2 is greater than the side

opposite 1, we know that m 1 is less than m 2

and m 5. Now, we have m 1 < (m 2 = m 5) < m 4 <

m 6. From the triangles, m 1 + m 2 + m 3 = 180 and

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

m 1 + m 3 must equal m 4 + m 6. Since m 1 is less than m 4, we know that m 3 is must

be greater than m 6. The side lengths list in order from smallest to largest is :m 1, m 2 = m 5, m 4, m 6, m 3.

48. WRITING IN MATH Explain why the hypotenuseof a right triangle is always the longest side of the triangle.

SOLUTION: Based on the Triangle Sum Theorem, we know that the sum of all the angles of a triangle add up to 180 degrees. If one of the angles is 90 degrees, then the sum of the other two angles must equal

degrees. If two angles add up to 90 degrees, and neither is 0 degrees, then they each must equal less than 90 degrees, making the sides across from these two acute angles shorter than the side across from the 90 degree angle, according to Theorem 5.10. Because the hypotenuse is across from the 90 degree angle, the largest angle, then it must be the longest side of the triangle.

49. STATISTICS The chart shows the number and types of DVDs sold at three stores.

According to the information in the chart, which of these statements is true? A The mean number of DVDs sold per store was 56. B Store 1 sold twice as many action and horror films as store 3 sold of science fiction. C Store 2 sold fewer comedy and science fiction than store 3 sold. D The mean number of science fiction DVDs sold per store was 46.

SOLUTION:

Mean of science fiction DVDs is or

46. The correct choice is D.

50. Two angles of a triangle have measures and .What type of triangle is it? F obtuse scalene G obtuse isosceles H acute scalene J acute isosceles

SOLUTION: By Triangle Sum Theorem, the measure of the third

angle is or

Since no angle measures are equal, it is a scalene triangle. And one of the angles is obtuse; so, it is an Obtuse Scalene Triangle. The correct choice is F.

51. EXTENDED RESPONSE At a five-star restaurant, a waiter earns a total of t dollars for working h hours in which he receives $198 in tips and makes $2.50 per hour. a. Write an equation to represent the total amount of money the waiter earns. b. If the waiter earned a total of $213, how many hours did he work? c. If the waiter earned $150 in tips and worked for 12 hours, what is the total amount of money he earned?

SOLUTION: a. The equation that represents the total amount of money the waiter earns is:

b. Substitute t = 213 in the equation and find h.

So, the waiter worked for 6 hours. c.

The total amount he earned is $180.

52. SAT/ACT Which expression has the least value? A B C D E

SOLUTION:

|–99| = 99

has the least value. So, the correct choice is E.

In , P is the centroid, KP = 3, and XJ = 8.Find each length.

53. XK

SOLUTION:

Here, .

By Centroid Theorem,

54. YJ

SOLUTION:

Since J is the midpoint of XY,

COORDINATE GEOMETRY Write an equation in slope-intercept form for the perpendicular bisector of the segment with the given endpoints. Justify your answer.

55. D(–2, 4) and E(3, 5)

SOLUTION:

The slope of the segment DE is or So, the

slope of the perpendicular bisector is –5. The perpendicular bisector passes through the mid point of the segment DE. The midpoint of DE is

The slope-intercept form for the equation of the perpendicular bisector of DE is:

56. D(–2,–4) and E(2, 1)

SOLUTION:

slope of the perpendicular bisector is

The perpendicular bisector passes through the mid point of the segment DE. The midpoint of DE is

57. JETS The United States Navy Flight Demonstration Squadron, the Blue Angels, flies in a formation that can be viewed as two triangles with a common side. Write a two-column proof to prove that

if T is the midpoint of

SOLUTION: The given information in this proof can help make congruent corresponding sides for the two triangles.

If T is the midpoint of , then it would form two

congruent segments, and . Using this, along with the other given and the Reflexive Property for the third pair of sides, you can prove these triangles are congruent using SSS.

Given: T is the midpoint of

Prove:

Proof: Statements(Reasons)

1. T is the midpoint of (Given)

2. (Definition of midpoint)

3. (Given)

4. (Reflexive Property)

5. (SSS)

Determine whether each equation is true or false if x = 8, y = 2, and z = 3.

SOLUTION:

The equation is false for these values.

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

eSolutions Manual - Powered by Cognero Page 1

7-3 Inequalities in One Triangle

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

. By substitution,

SOLUTION:

Angle: J, K, L

SOLUTION:

m 5 and m 8 < m 5 .

, and .

> m 8 .

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

Angle:

SOLUTION:

degrees and the

degrees. Therefore,

. Angle:

degrees. Therefore,

24. 1, 5, 6

25. 2, 4, 6

26. 7, 4, 5

27. 3, 11, 12

28. 3, 9, 14

and m 14.

29. 8, 10, 11

m 10 and m 11.

30. ABD, BDA

SOLUTION:

31. BCF, CFB

SOLUTION:

32. BFD, BDF

SOLUTION:

33. DBF, BFD

SOLUTION:

34. SM, MR

SOLUTION:

, , since 60 < 70.

35. RP, MP

SOLUTION:

by Theorem 5.10, .

36. RQ, PQ

SOLUTION:

Theorem 5.10,

37. RM, RQ

. The side

m 1 > m 2.

39. A(–4, 6), B(–2, 1), C(5, 6)

40. X(–3, –2), Y(3, 2), Z(–3, –6)

SOLUTION: a.

SOLUTION:

m 4 + m 5 + m 6 = 180. Since m 2 = m 5,

SOLUTION:

angle is or

SOLUTION:

|–99| = 99

53. XK

SOLUTION:

Here, .

54. YJ

SOLUTION:

55. D(–2, 4) and E(3, 5)

SOLUTION:

56. D(–2,–4) and E(2, 1)

SOLUTION:

Prove:

3. (Given)

5. (SSS)

SOLUTION:

use the exterior angle inequality theorem to the exterior angle inequality theorem to list all of...

Documents